## Evolution of Woods-Saxon energy levels

The Woods-Saxon potential is

$\displaystyle H = \frac{V_0}{1+\exp^{\frac{r-R_0}{a_0}}} - LS \frac{V_{so} \exp^{\frac{r-R_{so}}{a_{so}}}}{(\exp^{\frac{r-R_{so}}{a_{so}}})^2}$

The LS coefficient is

$\displaystyle LS = \frac{1}{2} (J(J+1) - L(L+1) - \frac{1}{2}\frac{3}{2})$

So, this is the neutron single particle levels.

## Physics behind Woods-Saxon energy levels

In the last post, I hope I explained how to find the Woods-Saxon energy levels for a given parameters. I just searched the best fit Woods-Saxon parameters to best fit the neutron single-particle energy for  209Pb. This is a double magic nucleus, and the there is no large fragmentation for the neutron excitation energy, thus, the outermost neutron in this nucleus can be considered as a good single-particle.

The best fit parameters are

The rms difference is just 78 keV!! For each level, the difference is not more than 50 keV!

So, how can we interpret this fitting result? It means, the energy levels can be well explained by WS mean field.

Recall that, the mean field actually included a lot things, it is the effective single particle potential that a nucleon is feeling. When we pushing the rms value to be minium. We are actually finding the best mean field. And the difference is due to the residual interaction.

But what contained in the mean field and residual interaction? The total Hamiltonian is

$\displaystyle H = H_{N} + H_{NN} + H_{NNN} + ....$

the mean field approach is to add an artificial potential $V$

$\displaystyle H = (H_{N} + V) + (H_{NN} + H_{NNN} +... - V) = H_{m} + H_{R}$

such that $latex H_{R}$ is minimum.

And by fitting energy levels using Woods-Saxon energy levels, we are basically doing the same thing! And the mean field is explicitly containing 2-body force, 3N-force and so on. So, can we say “because the Woods-Saxon can explain the energy level very well, the tensor force or other force is insignificant” ? the answer is no. Because the Woods-Saxon potential explicitly contains tensor force and other force.

## Calculation of Woods-Saxon energy levels

Hello everyone, I have been very busy lately, working on many things at the same time. Writing experiment proposal, testing a new detector array, flighting with the noise. Coding a program to read digitizer data for real time. Update the analysis code for coming experiments. Update/making some calculation codes, one of them is the calculation of Woods-Saxon energy levels.

Using RK4 method, a 2ndary 1-D differential potential can be solved easily (with some effort).

The Woods-Saxon potential with spin-orbital coupling is:

$\displaystyle H = \frac{P^2}{2m} + V(r) + \frac{1}{r} \frac{d}{dr}V_{so}(r)\langle L\cdot S\rangle$

$\displaystyle V(r) = \frac{V_0 }{1+ \exp{\frac{r-R_0}{a_0}}}$

$\displaystyle V_{so}(r) = \frac{V_{so}}{1+\exp{\frac{r-R_{so}}{a_{so}}}}$

There are 6 parameters, 3 for the central term, 3 for the spin-orbital term.

And the equation is

$\displaystyle H\Psi(r, \theta, \phi) = E \Psi(r, \theta, \phi)$

Since the radial wavefunction has to be zero when it is far away, we can search for energy, such that the wavefunction is close to zero at a certain range. ( this is the only method I know how to practically implemented. I can imagine that, there may be some other method, transform the whole equation with a fixed r and directly solve the energy. Something like least-square method for linear system can be transformed into Matrix formalism and be directly solved. But I don’t know how)

The code is available at github (currently only for neutron).

I also have an excel to solve a single energy. Feel free to leave in comment and I will email it to you.

## Root mean squared radius of WS

The Woods-Saxon potential is

$f(r) = 1/(1+Exp(\frac{r-R}{a}))$

where $R=r_0 A^{1/3}$ is the half-maximum radius that $r_0$ is the reduced radius and $A$ is the nuclear mass number, and $a$ is the diffusiveness parameter.

The root-mean-squared (rms) radius is

$\sqrt{\left} = \sqrt{\int{f(r) r^2 r^2 dr} / \int{f(r) r^2 dr}}$

where the “extract” $r^2$ is because of spherical coordinate.

The integration is a polynomial

$\int{f(r) r^n dr} = -a^{n+1} \Gamma(n+1) \sum_{k=1}^{\infty} \frac{(-Exp(R/a))^k}{k^{n+1}}$

In mathematica, the sum is notated by,

$PolyLog(n, x) = \sum_{k=1}^{\infty} \frac{x^k}{k^n}$

Thus, the rms radius for $a > 0$ is

$\sqrt{\left} = a \sqrt{12 \frac{PolyLog(5, -Exp(R/a))}{PolyLog(3, -Exp(R/a))} }$

For $a = 0$, $f(r) = 1, r < R$,

$\sqrt{\left} = \sqrt{\frac{3}{5}}R$

## Woods-Saxon Shape

this is a collective model of the nuclei density vs radius. it has another name Fermi-shape.

$\rho(r) = \frac { \rho_0} {1+ Exp( (r - R_0)/a) }$

where $\rho_0$ is central density, or density at r = 0. $R_o$ is the radius of half density and $a$ is the diffuseness. when a is large, the “tail” of the shape will be longer.

the radius is measured in unit of fm $1 fm = 10^{-15} m$.

$R_0$ can be vary from 1 fm to 7 or 8 fm.  for 16O, it is about 2 fm. and for 208Pb, it is about 6fm .

$a$ is more or less the same for different nuclei.

the density of nuclei can be mass density or charge density, the Woods-Saxon also gives a good approximation.

for mass density, a $10^{17} kg m^{-3}$. for comparison, water density is 1 kg per meter cube.

and charge density is about $0.15 e fm^{-3}$.